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Full text of "Analysis of prestressed and reinforced concrete folded plates"

ANALYSIS OF PRESTRi^ooED AND REINFORCED 
CONCRETE FOLDED PLATES 



by 



RAMESH GAMI 
B. S., Bombay University, 1958 



A MASTER'S REPORT 

submitted in partial fulfillment of the 
requirements for the degree 

MASTER OF SCIENCE 

Department of Civil Engineering 

KANSAS STATE UNIVERSITY 
Manhattan, Kansas 

1963 

Approved by: 




'J/.yn, 



ajor Professor 




k f ii 



TABLE OF CONTENTS 
r>->e.i+s Page 

SYNOPSIS 1 

INTRODUCTION 2 

DEFINITIONS 4 

STATEMENT OF PROBLEM 6 

ANALYSIS OF FOLDED PLATES CONSIDERING JOINT DISPLACE- 
MENT 7 

ANALYSIS 8 

Primary Stresses 8 

Secondary Stresses 20 

ANALYSIS OF POST SSB8X0X5S rRESTREoSED COHCBEH FOLDED 

PLATE 38 

CONCLUSIONS 55 

ACKNOWLEDGMENT 57 

APPENDIX I - EXPLANATION OF TERMS 58 

APPENDIX II - BIBTIOGRAPHT 60 



ANALYSIS OF PRESTRESSED AND 

REINFORCED CONCRETE 

FOLDED PLATES 



by 



RAMESH GAUI 1 



SYNOPSIS 



The simplified methods of analysis for long span pre- 
stressed and reinforced concrete folded plate structures are 
presented herein. 

A reinforced concrete folded plate is analyzed by resolv- 
ing the applied load into two directions, one vertical and the 
other parallel to the plate on which it acts. The vertical 
load will induce "slab" action and the parallel load will in- 
duce "plate" action. Slab action can be analyzed by assuming 
non-yielding supports and then applying a correction for de- 
flection, from which total plate loads are found. "Plate" ac- 
tion causes a deflection of the longitudinal edge and intro- 
duces longitudinal stresses. To satisfy the compatibility 
condition, a stress distribution is done. Moments are com- 
puted due to relative displacements. From these moments, 
plate loads and longitudinal stresses are calculated. 



Graduate student, Department of Civil xungineering, 
Kansas otate University, Manhattan, Kansas. 



To analyze the prestressed plate, the stress distribution 
along the longitudinal axis is analyzed in the same manner as 
for any homogenous beam of rectangular section. 

Transverse bending in successive plates is analyzed by 
the conventional moment distribution method utilized for con- 
tinuous structures. 

A typical problem is outlined and the simplified method 
of analysis is discussed in this report. 



INTRODUCTION 

Folded plates, or as they are sometimes called, prismatic 
shells or hipped plates, provide a useful and economical meth- 
od of construction for roof and floor systems in a wide variety 
of structures. They are competitive with other construction 
methods for short spans and have proven exceptionally economi- 
cal where relatively large spans are needed as for auditoriums, 
gymnasiums, industrial buildings, hangers, department stores 
and parking garages. The folded plate shape of roof structure 
has come into wide usage because of its low cost of construc- 
tion for long span, high load carrying capacity, rigidity, and 
aesthetic interest. Selection of concrete for the shell mate- 
rial furnishes a high degree of fire resistivity, ease of mold- 
ing to the desired alinement and profile, a great degree of 
permanence, and low construction and maintenance costs. 

Folded plate floor or roof construction consists of a 



series of repeated units, each of which is formed by tv/o or 
more flat plates intersecting at an angle. The plates act as 
a continuous slab transversely and as beams in their own planes. 
Their structural behaviour resembles that of shells. In fact, 
a cylindrical shell can be thought of as a folded plate in lim- 
it. The structural action of a folded plate consists of trans- 
verse "slab action" by which the loads are carried to the 
joints and longitudinal "plate action" by which the loads are 
finally transmitted to the transverses. Because of its great 
depth and small thickness, each plate offers considerable re- 
sistance to bending in its own plane. This "plate action" ex- 
plains the remarkable rigidity of folded plate construction. 

Folded plates have certain advantages over shells. These 
advantages are: 

(1) The shuttering required is relatively simpler as 
it involves only straight planks. 

(2) Shuttering can be stripped at the end of seven days, 
if not earlier, because of their greater rigidity; 
this results in quicker turnover which, in turn, 
cuts down construction time. 

(3) The design involves only simple calculations which 
do not call for a knowledge of higher mathematics. 

(4) Movable formwork can be employed for their construc- 
tion with greater ease than with cylindrical shells. 

(5) Simple rectangular diaphragms take the place of 
complicated transverses required for shells. 



(6) .?heir light reflecting geometry and pleasing out- 
lines make then comparable with shells in their aes- 
thetic appeal. 
Interest in and use of this type of roof has increased 
considerably in this country. Considerable additions have 
been made to our analytical and experimental knowledge in the 
last decade. The purpose of this report is to solve typical 
problems and discuss procedures which may be employed for the 
analysis of a single-span folded plate structure. 

DEFINITIONS 

The following definitions are used as a basis for the 
discussion in this report: 

(1) A plate is an individual planar element of the 
structure. 

(2) The length of a plate is the dimension between trans- 
verse supports. (Fig. 1, "L") 

(3) The width of a plate is the transverse dimention be- 
tween longitudinal edges. (Fig. 1, "W") 

(4) The height of the structure is the vertical dimension 
of the upper and lower extremes of a transverse cross 
section. (Fig. 1, "h") 




FIG. 1. - END PORTION OF A FOLDED PLATE ROOF 




Support 



Support 



FIG. 2. - TYPICAL CROSS SECTION THROUGH MULTIPLE FOLDS 



STATEMENT OF PROBLEM 

Consider a simple span of 60 feet in which we shall arbit- 
rarily assume a column spacing along the ends of this longer 
span of 23 feet. The slope of each plate will be assumed to be 
3 1/2 on 12, vertical to horizontal. (Fig. 2.) A trial analy- 
sis will indicate this corrugated configuration to be unsuit- 
able for conventionally reinforced concrete due to the very 
small area of concrete available at the ridge, causing high 
compressive stresses and a very high percentage of compressive 
steel reinforcement. This can be primarily attributed to the 
shallow ratio of depth from ridge to valley for the long span 
of 60 feet and the absence of sufficient concrete area in the 
ridge. 

Several remedies for this condition become readily appar- 
ent: 

(1) Steepen the slope of the plate and add a third plate 
horizontally at the ridge in order to provide suffi- 
cient concrete area to reduce the high compressive 
stresses. 

(2) Steepen the slope of the plates in order to provide 
a deeper section from ridge to valley. 

(3) Provide a combination of each of those stated in (1) 
and (2). 

(4-) Prestress the structure in order to utilize all of 

the concrete area as homogenous section. 
This report shall advance methods of analysis for design 



of folded plates by the above mentioned procedures (2) and (4). 

(a) To steepen the sloj^e of the plates in order to pro- 
vide a deeper section from ridge to valley. The folded plate 
is analyzed by considering the effects of relative displace- 
ments of the longitudinal edges. 

(b) To prestress the structure in order to utilize all 
of the concrete area as homogenous section. 

ANALYSIS OF FOLDED PLATES CONSIDERING 
JOINT DISPLACEMENT 

Assumptions 

The following general assumptions are made in analyzing a 
folded plate structure: 

1. The material is homogenous, uncracked and elastic. 

2. Longitudinal edge Joints are fully monolithic and con- 
tinuous; there is no relative rotation or translation 
of two adjoining plates at their common boundary. 

3. The principal of superposition holds, that is, the 
structure may be analyzed separately for the effects 
of its redundants and various external loadings and 
the results combined algebraically. 

4. Individual plates possess negligible torsional resis- 
tance and torsional stresses due to twisting of the 
plates can be neglected. 

5« The function of the supporting diaphragms or bents is 
to supply the end reactions for the plate action and 



8 



for the longitudinal slab action. 1'hey are assumed 
incapable of providing restraint against rotation of 
the ends of the plates in their own planes, but may 
provide some restraint for longitudinal slab bending. 

6. Longitudinal strain due to plate action varies linear- 
ly across the width of each plate (plane section re- 
mains plane). The rate of change of strain with re- 
spect to width ordinarily will differ from plate to 
plate from which it can be inferred that there will 

be some relative displacement of the joints of a cross 
section. 

7. Longitudinal slab action can be neglected; that is, 
slab bending carries the load applied to the surface 
of a plate to the longitudinal edges only, as in a one 
way slab. 

ANALYSIS 

Primary Stresses 

The folded plate for the structure, as discussed in 
"Statement of Problem", is deepened at ridge and valley as 
shown in the line diagram, Fig. 3» 




FIG. 3. - LIKE DIAGRAM OF FOLDED PLATE STRUCTURES. 



The data for this sample solution are as follows: 

Slab thickness = 4 1/2" 

Span between transverses * . . . 60' 0" 
Column spacing = 23' 0" 

Loading: 

Roofing 5 psf 

Snow load, insulation, and accoustatics ... 30 psf 
The loads on the inclined surfaces are as follows: 
For Plate AB (Fig. 4) 
Snow Load: 




FIG. 4. - PLATE AB. 



10 



cos* 



£732 



= 0.94-9 



sin* = 2 0.317 
5732 

Snow load on inclined surface 

= 30 x cos* = 30 x 0.949 = 29. 4-7 lbs/sq. in. 

Dead Load : 

Hoofing material = 5*00 lbs/sq. in. 

Slab 4.5 x 150 * 55.33 lbs/sq. in. 

12 

Total Load : 

Sum of above three 

= 29.^7 ♦ 5.00 + 55.33 

= 89.80 say 90.00 lbs/sq. in, 

Components of Total Load : ( Fig . 5 • ) 

This load can be resolved into normal and tangential 

components, Wn and Wt. 




FIG. 5. - NORMAL AND TANGENTIAL COMPONENTS ON PLATE AB. 



11 



Wn ■ W x cos <x 

= 90 x 0.949 

= 85.41 lbs/sq. in. 
Wt = W x sin <x 

= 90 x 0.317 

= 28.53 lbs/sq. in. 

For Plate BC (Jig. 6.) 




fc-j — 
FIG. 6. - PLATE BC. 



Snow Load: 



Dead Load: 



« as x 11.5 

12.92 
■ 27.62 lbs/sq. in. 



roofing ■ 5.00 lbs/sq. in. 

Dead load of slab m 55. 33 lbs/sq. in. 

Total Load : 

= 87.95 lbs/sq. ft. 

Components of Total Load : 

Normal component Vn ■ 87.95 x 11.5 

12.95 

= 77.5 lbs/sq. It. 

Tangential component = Vt = 87.95 x 6 

12.92 

= 40.60 lbs/sq. ft. 



12 



The load on DE is by observation the same as the load on £C, 
that is: 

w = 77.5 lbs./sq. ft. say 78 lbs./sq. ft. 

W t = 40.6 lbc/s^. ft. say 41 lbs./sq. ft. 

For the purpose of analysis, assume fictitious supports at 
joints B, G, D, and calculate the Lioments and reactions. It is 
assumed that the relative displacement of joints is not present. 
A transverse strip one foot wide is considered, treating it as 
continuous slab supported at the joints by non-yielding sup- 
ports. The Boment distribution for this condition is performed 
in Fig* 7* 







86#/i 


't 






76#/ft 




8( 


s# 


/t 


t 






'1 


' 


1 ■ 


\ 


• 


■ ' 


i 


f ) 


' 1 


r 


' 


\ 


' ' 


' 


r ^ 


1' ,r v | | , V I 1 j 


■ v i 1 i> v ■' ' » 1 


A 














i 


E 


r * 

c 


fD 















E 



A E 


c 







£ 




Dist. F. 


1 0.5 


0.5 1 







FEM +1725 


-1090 +1090 


-1090 +1090 






3,0. 


-635 -317.5 


+317.5 +625 






final n +1725 


-1725 +772.5 


-772.5 +1725 


-1725 



FIG. 7. - SLAB MOMENT DUE TO EXTERNAL LOAD. 



Reactions are as follows: 

Total reaction at B ■ 1124 lbs. 

Total reaction at C s 368 lbs. 

Total reaction at D = 1124 lbs. 



13 



These reactions were obtained on the assumption that fic- 
titious supports exist at B, C % and D and hence, on removal, 
they give e jual and opposite reactions. These reactions are 
placed on the plates as shown in fig* 8, 




FIG. 8. - PLATE SHOWING REACTIONS DUE TO SLAB ACTION. 

These plate loads due to the above reactions can be ob- 
tained as follows (see Fig. 9) J 



tan © x - § - 0.3333 

© x « 18° - 26» 
tan © 2 * jj-r = 0.521 

© 2 s 27° - 40' 

Considering joint B, (see Fig. 10) 

Comp. along; AB Comp. along BC 54-8 

sin(90 + © 1 + 9 2 ) = sin 90 s sin (ISO - © x 



*P 



Comp. along AB . ^ 8 x sin <*> » & 1 * V 

sin (180 - 57 - 9 2 ) 

= 5^8 x 0.96 = 526 lbs. 



14 




FIG. 9. - INCLINATION OF PLATES. 



Comp. along BC ■ 



54-8 x sin 90 
sin (180 - 57 - 9 2 ) 



548 
~ 0.?1 

= 770 lbs. 




5^8# 
FIG. 10. - COMPONENTS ALONG AB AND BC DUE TO REACTIONS 



Similarly due to the load of 576 lbs. the components along 

AB and BC can be obtained as follows: 

Comp. alonp; AB Comp. alom BC 576 

sin 90 = sin (90 * ^ 4 © 2 ; = sin (180 - 9 X - © 2 ; 

Comp. along AB = s JPf 1 g Q 3 ^ Q . Qg) 

* o?fr = 812 lbs# 



15 



576 x sin (90 f & 1 f © 2 ) 

Comp. along BC = sin (180 - ^ - © 2 ) 

= 576 cot (9« + w 2 ) 

- 576 x 0.96 
= 553 lbs. 

Considering joint C, seo Fig. 11. 




FIG. 11. - COMPONENTS DUE TC EEACT10H8 A? C. 



Comp. BC Coinp. along CD 

sin 54° - 40' " sin 90° 



Comp. along BC * 434 x ffiB g^L ~ 



sin 55° 

^ x 0.82 
= 297 lbs. 

Comp. along CD = s in 55*- 20 ' 

434 

" 0.82 

» 528 lbs. 
Due to 434 lbs. reaction 
Comp. along BC - 528 lbs. 
Conp. along CD = 297 lbs. 



434 

sin (180 f 2 9 2 ) 



40' 

"20^ 



16 



812 
316 



297 




FIG. 12. - TOTAL FORCES AT THE JOINTS 



The total forces at the joints are shown in Fig. 12. 



The total load along AB can be calculated as follows 




Load along plate AB =» 1358 * Wt x 6.32 

= 1358 f 29 x 6.32 
= 1521 lbs. 

Similarly total load along plate BC 




= 2149 f v/t x 12.92 
= 2149 f 528 
= 2677 Ids. 



17 



From these loadings we can find longitudinal stresses on 
each plate. In the analyses one treats each plate as a beam 
carrying loads and spanning between end diaphragms with no edge 
shear along joints. 



Plate AB (Fig. 13.) 
1521#/ft 





1 1 1 1 1 1 1 1 1 


a' 


so.o' 







B 



A 



rf***- 



B 






1 



1880#/in 2 




1880#/* 
FIG. 13. - STRESS DISTRIBUTION ALONG PLATE AB. 



f 


1 




M 


w x L 
8 




f 


p 

w x L x h 

s 


x 12 1 




8 x 2 x b 


x h 5 




2 T 2 

3 x w x L 





4 x b x h^ 

3 x 1521 x 60 2 x 12 

4 x 4.5 x 6.32 x 6.32 x 144 

■ 1880 lbs/sq. in. 

The stresses in top and bottom fiber are 

« 1880 lbs/sq. in. 

T t " t 

f ju • - 1880 lbs/sq. in. 



13 



Piatt BC. (Fig. 14.) 



2149#/ft 



B 



60.0' 



V 



4.5 



m^M 




FIG. 14. - STRESS DISTRIBUTION ALONG BC. 



-9ia#/ n a 



f * 



. 2 



X W Y. If 



4 x b x h 

x 2.49 x 60 x 60 x 12 



x 4.5 x 12.92 x 12.^2 
912 lbs/sq. in. 



1860 ^An 3 - 




6 ld&0#/in x 



912 #/in 



FIG. 15. - STRESSES AT JUNCTIONS OF PLATES 



19 



From the preceding analysis, it Is seen that the stresses 
at the junctions of plates AB and BC at B are different as 
shown in Fig. 15» but the compatibility condition requires them 
to be equal. To satisfy the compatibility condition, a stress 
distribution similar to the process of moment distribution is 
done where, 

Stiffness •< 



1_ 

bh 



Carry over factor ... = - # 

Since the thickness of AB and BC are equal, the stiffness fac- 
tors are as follows: 

K ab = £732 
= 0.160 

if L 

ab 12.92 
= 0.078 

The stress distribution is as follows: 





A 'A 


B % ( 


Z % 


D g 


E 


DF 


0.672 


0.328 0.5 


0.5 0.328 


0.672 


Stresses 


+1880 -1880 


-912 +912 


+912 -912 


-1880 


+1880 


CO. 


-325 +650 


-318 +159 


+159 -318 


+650 


-325 


Final 
Stress 


+1555 -1230 


-230 +1071 


+1071 -1230 


-1230 


+1555 



Sign Convention: 



Compression + V 



e 



Tention 



- V. 



e 



20 



Secondary stresses 

Assume that plates AB, BC, CD and BE are not rigidly 

jointed and solve for the free edge deflections. There will 

be stresses due to rotations of those plates which do not have 

a free edge. Considering a transverse strip one foot wide at 

the middle of the structure 
-5 x L 2 (f. - f K ) 



S * 



48 x E x h 
where f-. ■ stresses at left edge of plate 

f. = stresses at right edge of plate. 

The deflection of plate BC can be computed as follows: 

f f b 



S 



-5 x L 2 (f . - f. ) 



be 48 x £ x h 



5 x 60 x 60 1071 - (-1230) x 144 
48 x £ x 12.92 



5 x 5600 x 2300 x 144 
= 48 x 12.92 x E 

18 x 23 x 144 x lQ^ 
= 48 x 12.92 x E 



be = E 



S. , - 96.0jc 10 



Similarly for AB 

g - 5x 60 x 60 (-1230 - 155) x 144 
max = 48 x E x 6.32 

5 x 3600 x 2785 x 144 
■ ♦ 48 x 6.32 

S ab = ^ * 1Q 3 inches. 



21 



From these individual deflections, the composite deflec- 
tion is found out "by graphical construction as shown in Figure 
17. From Figure 17» we obtained the deflection of 



170 x 10"^ 
BC ■ — i — 6 . This deflection is obtained on the assumption 

that edges A, B, D, D, and E, are all free but actually this is 

not the case and the deflection will be very, very small due to 

the fixity of the Joints. Therefore, moments are calculated 

which are produced by this relative displacement. In case of a 

fixed end beam, moments induced at A and B by sinking of sup- 

ports B will be - — *— , see Fig. 16. 




FIG. 16. - MOMENT DUE TO DISPLACEMENT, 



Considering edge C as the fixed end and end B as the end 
free to rotate, the bending moment at C = • — *r- . 



22 



s A b " 2 ^¥^ ^ -" s 



e BC - 9 -H^ »» -i*- 




e BC : ^^^ ^ units 



Scale: 1" = 60" 



FIG. 17. - COMPOSITE DEFLECTION ALONG BC . 



I = 



b x t 



I = 



12 




1x4 


.? 


12 5 x 


12 


1 




19 x 


12 



M. 



3 x £ x 170 x 10- 



where 8 = 



19 x 12 x E x 12.92 x 12.92 

170 x lp5 

E 



L = 12.92 ft, 



hi - 1360 lb. ft. 



The calculation of shears is as follows 



23 



B 



12-92' 



& 



'at© 12.92' 



J 



1360 #•**■ 



*b K c 12.92 



= 105 lbs 



The components of the loads in the direction parallel to 
plates are 

Plate load along AB a ^§j 

= 148 lbs. 

Plate load along BC a 105 x 0.96 

= 101 lbs. 

Plate load along BC = 105 x 0.^6 

0.82 

= 71 lbs. 



24 



Plate load along BC is therefore = 101 - 71 

= 30 lbs. 
Plate loads along AB and BC are, (see Fig, 18.) 

Plate load along AB » 148 lbs. 

Plate load along BG ■ 30 lbs. 




FIG. 18. - PLATE LOADS ON AB AND BC, 



From this the moment on plate AB can be obtained as follows: 






B t E 



/ 



With tension as (-) and compression as ( + ) 

R x 60 = 2/3 x 60 x 148 x 30 and 
a 

M Q = 2/3 x 148 x 30 x 30 x 5/8 
= 55500 ft. lbs. 



For plate BG the moment is as follows: 



30# 



go' 



^L 



25 



M Q = 2/3 x 30 x 30 x $0 x 5/8 
= 75 x 150 
= 11250 ft. lbs. 



M 
Stress in plate AB » -x 

„ b x h 2 

Z * T — 



Z « 4 * 3 x 6 »?2 2 x 144 



108 x 40 

,3 



* 4320 in* 
Stress in plate AB - ^°? 2 * 12 

=» 154 psi 

Stress in plate BC ■ 8 

11250 x 12 
Z 

7 b x h 2 



4.5 x 12.92 x 12.92 x 144 
6 

Z « 18000 in 5 



Stress in plate BC = 112 ^ n ^ n 12 



18000 

- 7.5 psi 

Stress condition for plates AB and BC are as shown in 
Figure 19. 



26 




1.54- #/in 



154- #/i^ 



^7-5 #/<rf 



FIG. 19. - STRESSES IN AB AND BC. 

Compatibility requires stresses at the junction to be 
equal and hence stress distribution is carried out as follows: 



A 




3 


C 


0.672 


0.328 


-154 


+154 


-7.5 


+7.5 


+54.25 


-109.5 


+53 


-26.50 


-99.75 


+45.5 


+45.5 


-19.0 



cresses 
CO. 
Final Check 

The maximum deflection is obtained once again from which 
relative deflection is commuted by the graphical procedure as 
in i ? ig. 21. 

B ab = -5/48 x L 2 x (f t - f b ) 



= +5A8 x 60 x 



60 (4^5 + 99.7 5) 
E x 6.32 



■ 12 ' 2 $ * 10? inches 



be = -5/48 x 60 x 60 



(43.5 + 19.0) x 144 
12.92 x E 



5/48 x 3600 x 64.5 x 144 
12#92 



- 2 -?° J 1Q ^ inches 



27 



Individual deflections are as shown in Fig* 20, from which 
relative deflection at BC can be found out graphically as shown 
in . 21. Lfta 21, s bc = ll '° | 1Q inches 



We see that correction applied for a displacement of 

1 ? Q 1 1Q causes itself a displacement of 11,Q * 1Q . The 

11 x 10^ 
correction will have to apply for a displacement of *** w A 

which will be in form of geometric series, such as 

2 n 

a, ar, ar , ar 



Summation 



1 f r 



In the above case the series will be as shown below after 
taking out common factor of *jr 

170, 170 x £^» 170 x ( n *g^ 

170 



Summation = , . 11.0 

1 * T75 

m 170 x 170 

" 181 



160 



P^t^-K Q 160 X 10^ 

Correct S^ « 1 

Actual S, m = Actual Longitudinal stress Actual Moment 

oc - ■ f 

Initial S, Initial Longitudinal stress Initial Moment 
160.0 

= 0.94 



28 




FIG. 20. - INDIVIDUAL DEFLECTION AT AB AND BC . 



cj.-d = ^—y. in units 



r, 2.70 x 10? . 

b BC = — ' = m units 




o 11.0 x 10 ?. 

6-op = jg in units 



Scale: 1" = 10" 



FIG. 21. - RELATIVE DISPLACEMENT ALONG BC . 



29 



The longitudinal stresses at center line of structure due to 
deflection are multiplied by 0.94- by which we obtain the actual 
longitudinal stresses. 

Longitudinal stresses at the center line of the structure 
due to deflection are: 



Initial stresses 

Final stresses 
Initial xo 94 



-99.75 



-95.75 



^5.5 



+42.8 



45.5 



+42.3 



-19.0 



-17.8 



The total longitudinal stresses at center line are: 



A 


1 


C 


+1550.00 
-93.75 


-1230 
+42.80 


-1230 
+42.80 


+1071 
-17.80 


+1461.25 


-1187.20 


-1187.20 


+1053.20 



stresses due to load 
(psi) 

Stresses due to de- 
flections in psi 

Final stresses in psi 



Final end moments : 



A 




3 


_J 


* 


J 




+1725 


-1725 


+77-50 


-772.50 


+1725 










+1360 


-1360 













+1280 


-1280 






Final end moment due 
to loading 

'inal end moment due 
to deflection with- 
out correction 

Final end moment due 
to deflection with 
correction 
1360 x 0.94 * 1280 



30 



Moments at center of long n (60'), 



Moments due to load 



w x If 



c 78 x 12.92 x 12.92 
8 

= 1620 lb. ft. 

The moment diagram at center of long span is shown in 
Fig. 22. 




FIG. 22. - MOMENT AT CENTER OF LONG SPAN. 

Moment at quarter point of long span (60') due to deflection 

(Fig. 23): 

The moment due to deflection is assumed to have parabolic 

variation along the span of structure. 

nT) ,. , Q . 1280 x 13 x 13 
Ordinate ab = 30 x 30 — 

= 32u units 



Moment at quarter point of long span due to deflection = 
1280— 320 = 960 ft. lb. (See Fig. 23.) 



31 





>^ 






"TC 








-fi 








O 






.1 


u 


< *' 


* X5' 






15' 



1 — ar 



-w 

O 
oO 



FIG. 23. - MOMENT AT QUARTER POINT DUE TO DEFLECTION. 

Moment diagram at quarter point: (Fig. 24.) 
Moment at C * 772 + 1280 - 320 
* 1732 ft. lbs. 




FIG. 24. - MOMENT DIAGRAM AT QUARTER POINT OF LONG SPAN (60') 

Moment diagram at end of long span (60'): (?ig« 25.) 
Moment due to deflection ■ 




FIG. 25 - MOMENT DIAGRAM AT END, 



32 



Shear force at center line of long span: (Fig. 26.) 



86#/ft 



IU1 ! 1 1 4 1111 1 



2* 

1725^ 



^ 78#/ft 

B lll 11 1 11 11 i 1 1 1 1 ITT] c 

2050#f<t 



as.ss* 



FIG. 26. - SHEAR FORCE AT CENTER LINE OF LONG SPAN. 



For Span BC: 

Reactions due to Loadings: 

R B = R c = 78 x 2^2 = 505 lbs. 

Reactions due to Moment: 



R . R . 172^ -202? 
K B - *C 12.92 



* - 25.25 lbs. 



For Span AB: 



R B = 5^0 lbs. 



Total shear force at 

C = 505 - 25.25 ■ 479.75 lbs. 

B = 505 - 25.25 + 5^0 = 1011.75 lbs. 



33 



Shear force at quarter point of long span: (Fig. 27 •) 



86#/ft 

a 1 1 1 i i a 1 i i i 1 b 



78#/ft 



<o4-0# ' 505 * 505# 

FIG. 27. - SHEAR FORCE AT QUARTER POINT OF LONG SPAJN. 



For sjjan BC: 

Reactions due to loading 
R B = R c = 505 lbs. 

Reaction due to moment ■ negligible 
For span AB: 

R^ x 6.32 = 1725 + 86 x 6.32 x —^ 

R^ = 5-4-0 lbs. 

Total shear force at 
G = 505 lbs. 
B » 505 + 54-0 = 1045 lbs. 

Shear force at end of long span (60*): (Fig. 26.) 



86#/ft 



BPH 3 ^ilUUUI UUUlj C 772 Ft# 
: V^ 505.00 * i 



54-0* 73-60 #= 73. 6# 



FIG. 28. - SHEAR FORCE AT EHD. 



34 



For span BC: 

Reaction due to loading « 505 lbs, 

.Reaction due to moment = 1725 - 772 

12.92 

= 73.6 lbs. 

Total shear force at 

G = 505 + 73.6 = 578.6 lbs. 
B = 5^0 + 578.6 = 1118.6 lbs. 

Final results may be summarized as follows: 

Maximum longitudinal stresses (consider compression as 
positive and tension as negative) : 



, 


B 


G 


f 1461. 50 
psi 


-1187.20 
psi 


+1053.20 
psi 



Maximum moments (moments in ft. lbs.): 



B 


C 


D 


Location 


-1725 


-2052 


-1725 


fe mid span 


-1725 


-1732 


-1725 


@ 1/4& span 


-1725 


-772 


-1725 


at end 



35 



B 



6- 32 



12-92' 



t 



B.M. Diagram 
is plotted on 
Tension Side. 



PLATES AB AND BC. 




479.75 



SHEAR FORCE AT CENTER OF LONG SPAN. 



7 




2052 



BENDING MOMENT AT CENTER OF LONG SPAN 



FIG. 29. - BENDiNG MOMENT AND SHEAR FORCE DIAGRAMS 
AT CENTER OF LONG SPAN. 





B 


<t 


< 6-32' J 


12.92' 






w 



PLATES AB AND BC . 



36 




505 



SHEAR FORCE AT QUARTER POINT 




1732 



BENDING MOMENT AT QUARTER POINT. 



FIG. 30. - SHEAR FORCE AND BENDING MOMENT AT QUARTER POINT 



B 



37 



I- 



6- W 



->*- 



12.92' 



PLATES AB AND BC . 




578.6 



SHEAR FORCE AT END OF SPAN. 




772 



BENDING MOMENT AT END OF SPAN. 



FIG. 31. - BENDING MOMENT AND SHEAR FORCE AT END OF SPAN, 



t>* 



. IS OP POSMJiM I0B20 FKEST8E88£I) 

CONCRETE FOLDED PLATE 

x'he stress distribution acting axiaily ^long the longitu- 
dinal axis is the same for any homogeneous beam of rectangular 
section. The top surface will be in compression, the bottom 
in tension under gravity loads. Application of a sufficient 
pre stressing force at a suitable distance below the center of 
gravity of the section c ; in eliminate the tension stress. 

The transverse direction of the plate is analyzed as a 
continuous slab of length equal to the width of the plate and 
supported at the folded lines of the ridges and valleys. These 
spans being quite short thus require only nominal steel rein- 
forcing as dictated by slab thickness and bending moment. Re- 
ductions of the plate action deflections along the ridges, and 
valleys, due to application of prestressing force, permit anal- 
ysis of the transverse slab by moment distribution with a rea- 
sonable degree of accuracy. Distribution bars are provided in 
the longitudinal direction in an amount ordinarily utilized for 
temperature steel. 

The design of the structure is more easily understood with 
the assumption of one individual leaf isolated from the others. 
One may consider the plate oriented in its working position and 
compute its sectional properties about the horizontal axis 



"Long Span Prestressed Concrete Folded Plate Roofs", by 
J. Brough and B. Stephens, Proc. Am. Soc. G. E. , Jan., I960, 
p. 95. 



39 



extending through its center of gravity. Figure 32 indicates 
the terms used in the following presentation. 




FIG. 32. - SECTION PROPERTIES OF ONE FOLDED PLATE. 



First a slab thickness is assumed and then the moment of 
inertia of the cross section for the inclined position to the 
vertical axis is computed. The moment of inertia of one plate 
is derived as follows with a slab thickness of 4# inches, nor- 
mal to the slope, and with the slope of the plate being 3»5 
vertical to 12 horizontal. 
Thus, in Fig. 32: 
dA - tdx 
y « tanc£ 

fe-fc* 



dx 



12 dy 

-frr 

♦H/2 
-H/2 
♦H/2 
-H/2 



y 2 t dx 



_2+. 12 dv 



40 



U-l 1^2 fH/2 

-H/2 



T3~T 



12 t 

" " '■ ' .Li 



3 3 



12 t / H v 

T3 ( 57 + ^ 



H' t 



♦H/2 
-H/2 



H 5 t 



where t is vertical thickness of the slab 

i is the slope rise on 12 which is 3.5 

H is height of valley to ridge center to center 

t' is the •!«% thickness normal to slope 



Thus t - |— .• Ui x 



P P 
12^ + 3.5 



* 4,5/0.98 



4.6875 in. 



j . ft^p x (z.o.25) 5 



I - 87300 in 

The tensioning force required for zero stress in the bottom 
under working load, at midspan is 



I = 



e + k. 



41 



Where F is the prestress force 

e il the ecentricity of prestress force from the center 
of gravity. 

k is the kern point 
The top fiber stress, at raidspar under workin- load is 

f FH 
ct ^ A c°b 

Where f is the allowable concrete unit stress. 
c 

H is the Ytrtieal height of the section valley to ridge 

A^ is the area of concrete, 
c 

The bottom fiber stress at initial condition at tr isfer is 



f =* F i 

cb A 

A c 



1 + e - U fx /F ± 



k t 



Where subscript i indicates the condition at transfer of pre- 
stress. 

Area of one plate = 4.6875 x 11. 5 x 12 

= 647 in 2 



r A 647 L ^ 



r r Htt m 40.25 + 4.6875 oo n.n u 

r 2 

k. * k, = •—- = 6.0 in. 
t b c 



Where k. is top kern distance 

k. is bottom kern distance 

r is the radius of gyration 



42 



If one assumes the center of gravity of the tendons as 
being e' above bottom of section at midspan, the prestress 
force eccentric arm is 

e ■ c - e* = 16.4-7 i'-i. 

Assume e' ■ 6.0 in. 

The longitudinal bending moments acting on the section for 
the dead load of the slab, the applied dead loads such as the 
roofing, insulation and ceiling, and the live load for which 
the structure shall be designed are computed as follows: 

Dead load for slab 



&*S 12 144 " ^ K? on inclined surface 
Vertical load 

2Z— a ZS- at S6 PS 

cos<$> 079^ jK> '^ 

Total load per plate 

11.25 x 56.25 = 646 PSF. 

Additional applied dead load: 

Roofing (five ply built up asphalt and felt) * 5*0 FSF 
Concrete insulation and }4 in. acoustical plaster = 5»0 PSF 

Total applied dead load: 
10 x cos <J> ■ 9.6 

Applied dead load per plate 
9.5 x 11.5 = HI lbs. 

Snow load per plate 

25 x 11.5 ■ 287.5 lbs. 



43 



Moments can be computed as follows: 

M^ slab - 0.6*6 X *jp • 291*0 ft-kip 

2 
M dl a PP lied - 0.111 x ^|- * 50.0 ft-kip 

2 
HL, * .2875 x S|- « 129.6 ft-kip 

M tl " 47 °* 6 ft ~ ki P 

The required prestre3sing force to furnish zero stress in 
the bottom -\t mid span with total load moment is 



M ti 



470.6 x 12000 , n ft , . a 
88 16.47 + 6.0 = 250 '° klps 



■ e + K t 16.47 

F o " $k * ** kipS 

P * Total effective prestress after deducting losses. 
F ■ Total prestress just after transfer, using say 4 
cables, requires F = 250/4 «= 62.5 kips per cable. 
F. ■ 73^5 kips per tendon. 

Initial tensioning force required for each cable should 

increase adequately for overcoming tendon friction and wobble 

of conduit. 

F H 
Unit stress at mid span = F. = -r—g 

c b 

250 x 44.94 x 1000 
= 647 x 22.47 

= 770 vs i top comp. 

= psi bottom comp. 



44 



Under the slab load only, the unit stresses at time of 
transfer or prestress force will be 






i • - K 



¥ 



16.47 - 291. J 
57o 



x 2.54 x 1000 
= 1155 psi comp. 

Using allowable steel and concrete stresses specified by 

the Joint Committee 323 ACI, ASCE for pres tressed concrete for 

2 
tensioned member. 

Prestressing steel f 's « 240,000 psi 

f 's yield = 210,000 psi 



s 



ultimate strength of steel 



f ' at design load ■ 3750 psi 
c 

determine the area of prestress steel 

A s = f~ s lffioOO * *»73* ■*• in "» sa y 1 -75 sq. in. 
s 

Check the unit shear stress: 
V = hi | fiS « 33.3 k ips 

Assume tendons placed in parabolic curve with c-k = e 
» 22.47 - 6.0 - 16.47 in above bottom at ends 



Joint Committee $23 ACI, AJ3CS, for crestrsssed concrete 
for tensioned members. 



45 







FIG. 33. - LOCATION OF FRESTRESS FORCE ARRANGED IN 
PARABOLIC CURVE. 

Shear at cracking load in steel: 

xr 4 x F x h 
V s ' L 

Where V is shear in steel 
s 

F is prestressing force after losses 
L is length of span, center to center of supports; 61.5' 
h is vertical distance below center of gravity of con- 
crete section, 10.45 in. 

v 4 x 286 x 10.45 
V s * 61.5 x 12 

* 142 kips 

Shear force in concrete: 

V Q 
v„ = c^ 

C f 



Where Q is statical moment about center of gravity of concrete 
section 



46 



I is moment of inertia of section 

b is horizontal projection of i>late width 

19.1 x 23 60 loc; mm4 
v c = 87300 x 43 " 125 pSi 

The principal tensile 3 tress is gi;-en by 



s t 



v c 2 ♦ (t c /2f- - f 0/2 



Where f is allowable concrete unit stress 
c 



f c ' ^T 1 = 386 psi 



s t = ( >5 2 ♦ 193 2 

« 37 psi, which is satisfactory, since the allowable 
tensile stress is 110 psi. 

Check shear at ultimate load; 

u 13- 

Where V is ultimate concrete shear 
c 

2V - v g = 66.6 - 14.2 * 52.4 kips 

v is ultimate shear stress 

52.4 x 2560 
v u " 87300 x 4.5 

* 351 psi 



46 a 



Ultimate principal tensile stress is given by 



V 



v u 2 +< f c/2> 2 * f c/2 



551 2 +193 2 - 193 



» 199 psi, which is satisfactory, since the allowable 
tensile stress is 300 psi, thus design is satisfac- 
tory. 

Check section for cracking moment: 

The factor of safety against cracking moment 
* M/M g 

Where M - M Q + M„ 
s c 

M 8 = F (e + k t ) 

286 (16.47 ♦ 6.0) 
12 

= 514.10 ft. kip. 



M F_.I 0.14 f • x I 
r^ ■ c * c 
c 



% 



« c = °- 14 f 2 ^° x x il 7?0 ° ■ w "• "*• 



The factor of safety against cracking: 



514.10 - 170 , y. 
525 i# ^ 



The factor of safety against live load cracking: 



M - M 



; dl 



M 



11 



47 



Where H • M. ♦ ML ■ 684.10 ft. kip. 
s c 

^dl * s total dea( * load moment » 341.0 ft. kip. 
M,, is live load moment . . . - 129.6 ft. kip. 

The factor of safety against live load cracking: 

684.10 - 341.0 
= 120.6 — 

= 2.65 

The factor of safety considering ultimate moment: 
The factor of safety for total load 

"u 

^n~ ~~ -m ultimate tensile force x ultimate lever arm 
Where fi * pj 

s 

Ultimate tensile force ■ f ' . A a 

c s 

* 240,000 x 1.75 x 0.001 

m 420 kips. 

Ultimate lever arm can be computed as below: (See Fig. 34.) 
Ultimate lever arm m 44.94 - 6.0 - distance of center of 

concrete resistance below top 

Center of concrete resistance if found as below: (See Fig. 34-.) 
Equivalent width of top section « 12 * 4 *5 = 15.4-3 in. 
Area of top triangle = ^* 6< ? * 1 ^' 4 ? « 36.2 sq.in. 



48 




FIO. 34. - ULTIMATE MOMENT STRESS CONDITION. 



Additional compressive area of stem necessary is 

m Total area under compression - area of top triangle 
» 420 x 1000 



5190 
* 95. 3 s<i« in 



- 36.2 



Additional vertical depth of stem required is 

■ fill? = 6 - 2 in - 

Taking moment about top for finding center of concrete 

resistance "C ''. 
cr 

c m ?6.2 x 2/3 x 4.6? jfcjj (6.2/2+4.6? ) 
^cr 131.5 

■ b.5 in. below top 

Ultimate lever arm is 

« 44.94 - 6.0 - 6.5 

* 32.44 in. 
Ultimate moment 

- " 20 i^ 2 - m . 113 ft. kip. 



49 



The factor of safety for total load 
M s 11^8 

m O 

The factor of safety for live lord 

- M ; M di 

1138 - 341 
* 129.6 

= 6.15 

Transverse bending in successive plates is analyzed by 
the conventional moment distribution method utilized for con- 
tinuous structures. The valley and ridge serve as support for 
the slab. This being a short spsji, 11" - 6" f moments will be 
no greater than would be encountered in a level slab with sup- 
ports at the same frequency. 

Analysis of a 1 ft. wide strip transverse to the main span: 
W dl * 54 ♦ 10 

= 64 lbs. 
w\ • » 25 cos (J) 

= 24.5 lbs. 
V total load =64+24.5 
= 88.5 lbs. 
Assuming a strip of unit width of cross sectior of the 
structure to act as a continuous one way slab on unyielding 



50 



supports, the ridge moments are determined by the moment dis- 
tribution. Because of symmetry, the slab may be assumed as 
rigidly fixed at ridge. 




11.5* 



A 



B 






1 


5/7 


V7 


+1725 


-1060 
-665 


+1060 
-333 


-1060 


+1060 






+143 


+190 


+95 


+1725 


-1725 


+870 


-870 


+1155 



FIG. 35. - RIDGE MOMENTS. 



Ridge moments are computed in Figure 35* 

Maximum bending moment at the center of span occurs when 
that span and every other alternate span is loaded. Thus, bend- 
ing moment at center due to uniformly distributed loading can be 



^"Design of Folded Plates", by Eliahu Tranm, Proc. Am. Soc. 
C.E., Vol. 85, Oct., 1959, p. 87. 



51 



obtained by the following formula: 

riidsj;an positive bending moment ■ rr- (w, + V, /p) L 

= }g x (64 ♦ 12.25) x 12 2 

= 912 lb. ft. 

Distribution bars are i>rovided in the longitudinal direction 
in an amount ordinarily utilized for temperature steel. The 
positive moment reinforcement, extending transversely, was 
placed on top of the distribution steel. The prestressing ten- 
dons were placed directly on top of the bottom steel and tied 
to the bottom mat at the correct location. 

Lfter all tendons were placed, the end bearing plates were 
substantially anchored to the end form. 

Reinforcing bars for slab negative moments at valleys and 
ridges, stirrups, ties and grids for resistance of bursting of 
concrete at the end anchorage were placed last. 
Deflection: 
The moment due to pres tress is 

P x e = 294 x jo ■ W kip-ft for uniform loading 
plus 294 x p ' * 256 kip-ft for parabolic prestress. 

K dl slab = 29.10 ft-kip 

M dl a P^ lied ■ 50.0 kip-ft. 
n i;L = 150.6 ft-kip. 



52 



Upward deflection due to uniform prestress of 14-7 kip-ft: 
o 



W x IT 



147 x 6Q 2 x 12 2 x 12000 
8 x 3,500,000 x 87300 



0.374 in. 



Upward deflection due to parabolic prestress is 
5 x V x L 2 

* Sth — 

5 x ?56 x 60 2 x 12 2 x 12000 
= 3,500,000 x 87300 

« 0.5^3 in. 



Total instantaneous upward deflection due to prestress 

- 0.374 + 0.543 

* 0.917 in. 

Deflection after losses 
» 0.85 x 0.917 

• 0.78 in. 

Downward deflection due to slab load 

5 x 291 x 60 2 x 12 2 x 12000 
48 x 3,500,000 x 87300 

= 0.617 in. 

H«n04 immediate upward deflection at transfer is 

0.917 - 0.617 

- 0.300 



53 



Dead load deflection downward 

30 x 0.617 
* 391.0 

« 0.162 in. 



Net downward deflection after losses of prestress and effect 
of creep is 

» 0.300 - 0.162 

« 0.138 in. 

Instantaneous downward deflection due to live load moment is 

130.6 x 0*617 
291.0 

- 0.275 in. 

The resulting camber due to application of prestressing force 
supports the plate in its geometrical position thus eliminating 
critical secondary stresses caused by rotation of the ends at 
the supports that would be inevitable with deflections encoun- 
tered in conventionally reinforced concrete. 

The ends of the plates behind the anchorage are thickened 
to provide for distribution of the concentrated force applied 
by the tendoms. Additional reinforcing in the form of trans- 
verse and tie steel is provided in the area immediately behind 

the tendon anchorages to resist the bursting of the concrete 

A, 
created by the ores tress force. 

The end leaves of a transverse section through the 



4 

"Long span prestressed concrete folded platf* roofs", by 

J. Brough and B. Stephens, Proc. km. 3oc. 0. *•« Jan. I960, 
Vol. 85, p. 91. 



5* 



structure must be provided with tension ties to resist the 
horizontal force component. This may be accomplished by pro- 
viding a tie beam, gable wall or other means, at or near the 
support, that is adequate for resistance of horizontal reac- 
tion. 



55 



CONCLUSIONS 



I'he first structures of this kind were large coal bunkers 
designed and erected by G. Ehlers of Germany in 1924—25. The 
first paper on the subject was published by him in 1930. His 
analysis assumed the longitudinal joints to be hinged, neglect- 
ing the transverse moments at the junction of the plates. The 
displacement of joints was also ignored. This theory was im- 
proved upon in 1932 by £• Gruber who included the effects of 
transverse continuity and joint displacements. Assuming the 
joints to be hinged as a first approximation, he developed a 
solution in the form of simultaneous differential equations of 
the fourth order, which were solved by the use of rapidly con- 
verging series. This approach involves (7n + 2) unknowns for 
(n + 1) plates; thus, a roof of 5 plates would involve 30 un- 
knowns. Although solution proposed by Gruber was very labori- 
ous, his conclusion that the assumption of hinged or rigid 
joints would considerably affect the final results was signifi- 
cant. This work was followed by that of Cramer who published 
a paper in 1953. He laid down rough limits in terms of the 
length bo width ratio of individual plates for their classifi- 
cation as "long" and "short". The paper by Winter and Pei pub- 
lished in 194-7 is a landmark in the theory on che subject as 
thi,/, for the first time, reduced the algebraic solution into 
a stress distribution procedure analogous to the well-known mo- 
ment distribution method. However, they neglected the displace* 
ment of the joints. For short folded plates their approach 



56 



offers a very simple desig,_i procedure. However, for long 
plates the joint deflections cannot be ignored, Girkmann, in 
his book published in 194-8, takes into account joint displace- 
ments. Treating transverse moments at the joints as the un- 
knowns, he formulated conditions for the compatibility of lon- 
gitudinal stresses and displacements at joints. The method 
leads to as many simultaneous equations as the unknown trans- 
verse moments. The paper presented by Whitney at the joint 
ASCE-IABSE meeting in New York is a presentation in English of 
the Girkmann method with some modification. Gaafar in 1953 
published a modification of the Winter and Pie method extended 
to include the effect of joint displacements. 

Among available methods, the winter and i ; ie procedure is 
the simplest. It is applicable only to short folded plates for 
which the joint displacements can be ignored without appreciable 
error. Of the methods that are applicable to folded plates of 
all proportions, those due to Gaafar which is considered in this 
report seems to be most suitable. 



57 



ACKNOWLEDGMENT 

The writer wishes to express his sincere gratitude to 
Dr. John Mc Entyre for his kind guidance and assistance in 
the preparation of this report. 



58 



APPENDIX I - EXPLANATION OF TERMS 

a ■ Slope rise on 12 

A = Area of concrete 
c 

A ■ Area of steel 
e.g. = Center of gravity 
dl ■ Dead load 

e = Ecentricity of prestress force from the e.g. 
e' = Distance from surface to center of prestress force 

F ■ Prestress force after loboes 

F ■ Modulus of rupture of concrete 

F^ = Prestress force at tiiae of transfer 
o 

f = Allowable concrete unit stress 
c 

f ■ ■ Design concrete strength 

f ■ Allowable unit stress 
s 

H ■ Vertical height of section valley to ridge 

h * Vertical distance above or below e.g. of concrete 
section 

I ■ Moment of inertia of section 

i ■ Subscript indicating condition at transfer of prestress 

k = Kern point 
11 = Live load 

M m Bending moment 

Q ■ Statical moment about e.g. 

r = Radius of gyration 

S = Principal tensile stress 
S' = Principal tensile stress for ultimate 



59 



S. * Principal tensile stress 

S. , » Ultimate principal tensile stress 

c 

t = Thiclmess of slab normal to section 

t' * Thickness of slab vertically 

tl - Total load 

v = Unit 3 hear 

v » Unit shear stress in concrete 
c 

v = Ultimata shear stress 

V « Ultimate concrete shear 
c 



60 



APPENDIX II - BIBLIOGRAPHY 



"Design of Folded Plate Roofs," by Howard Simpson, 
Proc. Am. Soc. C. E. , Vol. 84, No. STI, January, 1958. 

"Hipped Plate Construction," by G. Winter and M. Pie, 
Jour. Am. Con. Inst., Proc. Vol. 43, January, 1947. 

"Reinforced Concrete Folded Plates Construction," by 
G. S. Whitney, B« G. Anderson, and N. Birnbaum, Jour. Struct. 
Div. , Proc. Am. Soc. C. £•, Vol. 84, January, 1958. 

"Hipped Plate Analysis, Considering Joint Displacements," 
by I. Gaafar, Tran. Am. Soc. C. £. , Vol. 119, 1954. 

"Design of Folded Plates," by Eliahu Traum, Proc. Am. 
Soc. C. X., Vol. 85, No. STI. 8., October, 1959. 

"Long Span Prestressed Concrete Folded Plate Roofs," by 
J. Brough and B. Stephens, Proc. Am. Soc. C. E. , Vol. 119, 
1954. 

"Design of Prestressed Concrete Structures," by T. Y. Lin. 

"Tentative Recommendations for Prestresaed Concrete," by 
ACI-ASCE Joint Committee 325. 

"Design and Calculation of Reinforced Concrete," by 
K. L. Rao. 

"The Analysis and Design of Folded Plates," by Ramswamy, 
Ramaiah and Jain, Indian Concrete Journal, Jul} , 1961. 



ANALYSIS OF PRESTRESSED AND REINFORCED 
CONCRETE FOLDED PLATES 



by 



RAMESH GAKI 
B. S., Bombay University, 1958 



AN ABSTRACT OF 
A MASTER'S REPORT 



submitted in partial fulfillment of the 
requirements for the degree 

MASTER OF SCIENCE 

Department of Civil Engineering 



KANSAS STATE UNIVERSITY 
Manhattan, Kansas 



1963 

Approved by: 




Mix* 



ajor Professor 




The simplified methods of analysis for lor.p. spun pro- 
stressed and reinforced concrete folded plr>te structures are 
presented herein, 

A reinforced concrete folded plate is analyzed b,/ solving 
the applied load into two directions, one vertical and the 
other parallel to the plate on which it acts. Vertical load 
will induce "slab" action and the parallel load will induce 
"plate" action, olab action can be analyzed by assuming non- 
yielding supports and then applying a correction for deflec- 
tion, from which total loads are found, 1 ,e" action causes 
a deflection of the longitudinal edge and introduces longitu- 
dinal stresr.es. To satisfy the compatibility condition a 
stress distribution is done, homents are computed due to rela- 
tive displacements. Prom these moments plate loads and longi- 
tudinal stresses are calculated. This process is repeatedly 
carried out and eve y time correction is applied. This correc- 
tion will be in form of geometric series, 3ummation of this 
series gives the actual deflection. The ratio of actual de- 
flection and initial deflection gives the multiplier which, 
when multiplied with initial lon ; ituainal stresses, ^ives ac- 
tual longitudinal stresses. The final longitudinal stress is 
calculated by adding stresses due to loading and stresses due 
to deflections. tmantl are calculated at quarter point, end, 
and at half distance of the longitudinal edge. 



To analyze the prestressed plate, the stress distribution 
along the longitudinal axis i3 analyzed in bhe same manner as 
for any homogenous beam of rectangular section. 

Transverse bending in successive plates is analyzed by 
the conventional moment distribution method utilized for con- 
tinuous structures. 

A typical problem is outlined and the simplified method 
o- analysis is discussed in this report.